The vapour in equilibrium with a pentane- hexane solution at 298K has a mole fraction of pentane equal 0.15 at 298K.what is the mole fraction of pentane in the solution.given vapour pressure of pure pentane and hexane are 511 torr and 150 torr respectively
Hi Lulu--Please check you work to make sure you have typed this problem correctly. I believe you must have more information; specifically, I think you need the total pressure of the vapor phase to complete the problem. Thanks.
Well,l there is a way with the information you posted.
Popentane = 511 mm
Pohexane = PoH = 150 mm
The total pressure of the vapor = Ptotal = Xpentane*511 + Xhexane*150
Xpentane in vapor = 0.15 = Xpentane/Ptotal = 511Xpentane/Ptotal so substitute Ptotal as follows and solve for Xpentane.
0.15 = 511*Xpentane/[Xpentane*511 + (1-Xpentane)*150] and solve for Xpentane. The answer is close to 0.05 mole fraction for pentane. Post your work if you get stuck.